Question

A $10\mu F$ capacitor and a $20\mu F$ capacitor are connected in series across a 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor?

Answer 1

Step 1: Calculating initial charge distribution

In series combination of two capacitors,
$C_{eq}=\frac{10\mu F\times 20\mu F}{10\mu F+20\mu F}$

$C_{eq}=\frac{20}{3}\mu F$

Since, two capacitors are connected in series charges on both of them will be same.

$Q=\left(C_{eq}\right(V)$
$Q=\frac{20}{3}\times {10}^{-6}F\times 200V$
$Q=\frac{4}{3}\times {10}^{-3}C$

Step 2: Calculating common potential

From charge conservation
Total initial charge = Total final charge

$Q+Q=q_1+q_2$
$2Q=(10\times {10}^{-6}F)V^{\prime }+(20\times {10}^{-6}F)V^{\prime }$
Where V' is common potential,
$V^{\prime }=\frac{2\times \frac{4}{3}\times {10}^{-3}C}{30\times {10}^{-6}F}=\frac{800}{9}Volts$
Thus, common potential is 800/9V

Hence, correct option is (B).